Question

我试图通过组变量拆分嵌套列表。请考虑以下结构：

> str(L1)
List of 2
 $ names:List of 2
  ..$ first: chr [1:5] "john" "lisa" "anna" "mike" ...
  ..$ last : chr [1:5] "johnsson" "larsson" "johnsson" "catell" ...
 $ stats:List of 2
  ..$ physical:List of 2
  .. ..$ age   : num [1:5] 14 22 53 23 31
  .. ..$ height: num [1:5] 165 176 179 182 191
  ..$ mental  :List of 1
  .. ..$ iq: num [1:5] 102 104 99 87 121

现在我需要生成两个列表，它们使用L1$names$last进行拼接，生成L2和L3，如下所示：

L2：按L1$names$last分组的结果

> str(L2) 
List of 3
 $ johnsson:List of 2
  ..$ names:List of 1
  .. ..$ first: chr [1:2] "john" "anna"
  ..$ stats:List of 2
  .. ..$ physical:List of 2
  .. .. ..$ age   : num [1:2] 14 53
  .. .. ..$ height: num [1:2] 165 179
  .. ..$ mental  :List of 1
  .. .. ..$ iq: num [1:2] 102 99
 $ larsson :List of 2
  ..$ names:List of 1
  .. ..$ first: chr [1:2] "lisa" "steven"
  ..$ stats:List of 2
  .. ..$ physical:List of 2
  .. .. ..$ age   : num [1:2] 22 31
  .. .. ..$ height: num [1:2] 176 191
  .. ..$ mental  :List of 1
  .. .. ..$ iq: num [1:2] 104 121
 $ catell  :List of 2
  ..$ names:List of 1
  .. ..$ first: chr "mike"
  ..$ stats:List of 2
  .. ..$ physical:List of 2
  .. .. ..$ age   : num 23
  .. .. ..$ height: num 182
  .. ..$ mental  :List of 1
  .. .. ..$ iq: num 87

L3：每个组只允许出现一次L1$names$last

List of 2
 $ 1:List of 2
  ..$ names:List of 2
  .. ..$ first: chr [1:3] "john" "lisa" "mike"
  .. ..$ last : chr [1:3] "johnsson" "larsson" "catell"
  ..$ stats:List of 2
  .. ..$ physical:List of 2
  .. .. ..$ age   : num [1:3] 14 22 23
  .. .. ..$ height: num [1:3] 165 176 182
  .. ..$ mental  :List of 1
  .. .. ..$ iq: num [1:3] 102 104 87
 $ 2:List of 2
  ..$ names:List of 2
  .. ..$ first: chr [1:2] "anna" "steven"
  .. ..$ last : chr [1:2] "johnsson" "larsson"
  ..$ stats:List of 2
  .. ..$ physical:List of 2
  .. .. ..$ age   : num [1:2] 53 31
  .. .. ..$ height: num [1:2] 179 191
  .. ..$ mental  :List of 1
  .. .. ..$ iq: num [1:2] 99 121

我试图应用this solution，但似乎这对嵌套列表不起作用。

可重现的代码：

L1 <- list("names" = list("first" = c("john","lisa","anna","mike","steven"),"last" = c("johnsson","larsson","johnsson","catell","larsson")),"stats" = list("physical" = list("age" = c(14,22,53,23,31), "height" = c(165,176,179,182,191)), "mental" = list("iq" = c(102,104,99,87,121))))

L2 <- list("johnsson" = list("names" = list("first" = c("john","anna")),"stats" = list("physical" = list("age" = c(14,53), "height" = c(165,179)), "mental" = list("iq" = c(102,99)))), "larsson" = list("names" = list("first" = c("lisa","steven")),"stats" = list("physical" = list("age" = c(22,31), "height" = c(176,191)), "mental" = list("iq" = c(104,121)))), "catell" = list("names" = list("first" = "mike"),"stats" = list("physical" = list("age" = 23, "height" = 182), "mental" = list("iq" = 87))))

L3 <- list("1" = list("names" = list("first" = c("john","lisa","mike"),"last" = c("johnsson","larsson","catell")),"stats" = list("physical" = list("age" = c(14,22,23), "height" = c(165,176,182)), "mental" = list("iq" = c(102,104,87)))), "2" = list("names" = list("first" = c("anna","steven"),"last" = c("johnsson","larsson")),"stats" = list("physical" = list("age" = c(53,31), "height" = c(179,191)), "mental" = list("iq" = c(99,121)))))

编辑：请注意，实际数据集非常大，并且比提供的示例嵌套得更深。

Answer 1

通常，对于修改列表，您需要使用递归。例如，考虑这个函数：

var triangleCoords = [
    {lat: A, lng: A},
    {lat: B, lng: B},
    {lat: C, lng: C}]

它需要一些列表foo <- function(x, idx) { if (is.list(x)) { return(lapply(x, foo, idx = idx)) } return(x[idx]) }和一些索引x。它将检查idx是否是一个列表，如果是这种情况，它将自己提供给列表的所有子元素。 x不再是列表后，我们会使用x给出的元素。在整个过程中，原始列表的结构将保持不变。

这是一个完整的例子。请注意，此代码假定列表中的所有向量都有5个元素。

idx

Answer 2

这不是一个完整的答案，但我希望它有所帮助。

看看这是否适用于L3：

x = data.frame(L1, stringsAsFactors = F)
y = x[order(x$names.last),]
y$seq = 1
y$seq = ifelse(y$names.last == shift(y$names.last),shift(y$seq)+1,1)
y$seq[1] = 1

z = list(list(names=list(first=z[[1]]$names.first, last=z[[1]]$names.last), stats=list(physical = list(age =z[[1]]$stats.physical.age, height= z[[1]]$stats.physical.height), mental=list(iq= z[[1]]$stats.iq))), list(names=list(first=z[[2]]$names.first, last=z[[2]]$names.last), stats=list(physical = list(age =z[[2]]$stats.physical.age, height= z[[2]]$stats.physical.height), mental=list(iq= z[[2]]$stats.iq))))

转换回列表的最后一部分（z）可以通过循环完成。假设相同的名称没有出现太多，循环不会太慢。

您说它更嵌套，在这种情况下，您需要添加is.null和/或tryCatch函数来处理错误。

拆分和操作嵌套列表

2 个答案: